Question : RRFRNDS I used brute force to solve this and was expecting a TLE but got a WA instead Somebody provide me some hints to solve this
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| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3856 |
| 2 | jiangly | 3747 |
| 3 | orzdevinwang | 3706 |
| 4 | jqdai0815 | 3682 |
| 5 | ksun48 | 3591 |
| 6 | gamegame | 3477 |
| 7 | Benq | 3468 |
| 8 | Radewoosh | 3462 |
| 9 | ecnerwala | 3451 |
| 10 | heuristica | 3431 |
| Страны | Города | Организации | Всё → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 1 | cry | 168 |
| 2 | -is-this-fft- | 162 |
| 3 | Dominater069 | 160 |
| 4 | Um_nik | 159 |
| 5 | atcoder_official | 156 |
| 6 | djm03178 | 153 |
| 6 | adamant | 153 |
| 8 | luogu_official | 149 |
| 9 | awoo | 148 |
| 10 | TheScrasse | 146 |
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Codeforces (c) Copyright 2010-2025 Михаил Мирзаянов
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Время на сервере: 17.02.2025 06:54:49 (k1).
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You can find editorial at codechef. Editorial link has to be under the statement as i remember.
Let's call list[i] is all friend of user[i], after that, we check all pair (i, j), if currently i and j is not friend and exists an user in list[i] is also friend of j, then pair (i, j) is valid, we increase the answer to 1.
That's an O(N^3) implementation and will time out